Optimal. Leaf size=21 \[ -\frac {3}{-e+4 \left (-\frac {5}{32 x^2}+x\right )+\log (4)} \]
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Rubi [A] time = 0.24, antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 5, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 1593, 6688, 12, 1588} \begin {gather*} \frac {24 x^2}{-32 x^3+8 x^2 (e-\log (4))+5} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 1588
Rule 1593
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {240 x+768 x^4}{25-320 x^3+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \left (e^2+\log ^2(4)\right )} \, dx\\ &=\int \frac {x \left (240+768 x^3\right )}{25-320 x^3+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \left (e^2+\log ^2(4)\right )} \, dx\\ &=\int \frac {48 x \left (5+16 x^3\right )}{\left (5-32 x^3+8 x^2 (e-\log (4))\right )^2} \, dx\\ &=48 \int \frac {x \left (5+16 x^3\right )}{\left (5-32 x^3+8 x^2 (e-\log (4))\right )^2} \, dx\\ &=\frac {24 x^2}{5-32 x^3+8 x^2 (e-\log (4))}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 27, normalized size = 1.29 \begin {gather*} -\frac {24 x^2}{-5-8 e x^2+32 x^3+8 x^2 \log (4)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 28, normalized size = 1.33 \begin {gather*} -\frac {24 \, x^{2}}{32 \, x^{3} - 8 \, x^{2} e + 16 \, x^{2} \log \relax (2) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 28, normalized size = 1.33
method | result | size |
risch | \(\frac {3 x^{2}}{x^{2} {\mathrm e}-2 x^{2} \ln \relax (2)-4 x^{3}+\frac {5}{8}}\) | \(28\) |
gosper | \(\frac {24 x^{2}}{8 x^{2} {\mathrm e}-16 x^{2} \ln \relax (2)-32 x^{3}+5}\) | \(29\) |
norman | \(\frac {24 x^{2}}{8 x^{2} {\mathrm e}-16 x^{2} \ln \relax (2)-32 x^{3}+5}\) | \(29\) |
default | \(\frac {3 \left (\munderset {\textit {\_R} =\RootOf \left (25+1024 \textit {\_Z}^{6}+\left (-512 \,{\mathrm e}+1024 \ln \relax (2)\right ) \textit {\_Z}^{5}+\left (-256 \,{\mathrm e} \ln \relax (2)+256 \ln \relax (2)^{2}+64 \,{\mathrm e}^{2}\right ) \textit {\_Z}^{4}-320 \textit {\_Z}^{3}+\left (80 \,{\mathrm e}-160 \ln \relax (2)\right ) \textit {\_Z}^{2}\right )}{\sum }\frac {\left (16 \textit {\_R}^{4}+5 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R}^{3} {\mathrm e}^{2}-32 \,{\mathrm e} \ln \relax (2) \textit {\_R}^{3}-80 \textit {\_R}^{4} {\mathrm e}+32 \textit {\_R}^{3} \ln \relax (2)^{2}+160 \textit {\_R}^{4} \ln \relax (2)+192 \textit {\_R}^{5}+5 \textit {\_R} \,{\mathrm e}-10 \ln \relax (2) \textit {\_R} -30 \textit {\_R}^{2}}\right )}{2}\) | \(144\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 26, normalized size = 1.24 \begin {gather*} -\frac {24 \, x^{2}}{32 \, x^{3} - 8 \, x^{2} {\left (e - 2 \, \log \relax (2)\right )} - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.39, size = 27, normalized size = 1.29 \begin {gather*} \frac {24\,x^2}{-32\,x^3+\left (8\,\mathrm {e}-16\,\ln \relax (2)\right )\,x^2+5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.50, size = 26, normalized size = 1.24 \begin {gather*} - \frac {24 x^{2}}{32 x^{3} + x^{2} \left (- 8 e + 16 \log {\relax (2 )}\right ) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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